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BMO Combinatorics problem

Adrian teaches a class of six pairs of twins. He wishes to set up teams for a quiz, but wants to avoid putting any pair of twins into the same team. Subject to this condition:
In how many ways can he split them into three teams of four?

Can anyone see how to do this? It's a simple problem when splitting into two teams of 6 (the precursor), but 3 teams of 4 becomes quite tough!
Sorry for the rough work, I was literally writing at the same time as thinking. Hopefully this solution makes sense, though I can't guarantee it's mistake free.IMG_0627.jpg
Original post by TimGB
Sorry for the rough work, I was literally writing at the same time as thinking. Hopefully this solution makes sense, though I can't guarantee it's mistake free.


And divide by 3!, giving 960, as the order of the teams doesn't matter.
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MaffsIsFun
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Original post by TimGB
Sorry for the rough work, I was literally writing at the same time as thinking. Hopefully this solution makes sense, though I can't guarantee it's mistake free.IMG_0627.jpg


Thank you that was really helpful - I got stuck by not realising that we had to use BOTH of the existing twin pairs in making Team 2, I guess there's a sort of pigeonhole principle-logic in that part. As the other commenter mentioned I think we need to divide by 3!, but other than that I'm pretty sure you've cracked it!
Original post by MaffsIsFun
...


Since you already have a solution, as an alternative method:

Spoiler

Original post by ghostwalker
And divide by 3!, giving 960, as the order of the teams doesn't matter.


Ah yes, I knew I'd forget something!

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